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Evgesh-ka [11]
4 years ago
8

Help plzzzz! pre calc question

Mathematics
1 answer:
Mumz [18]4 years ago
7 0

Answer:

\sum^{\infty}_0 {(-3)^n}

Step-by-step explanation:

Notice that the series: 1 - 3 + 9 - 27 +... clearly has powers of the factor 3 in its terms and it is also an alternate series (the terms alternate from positive to negative). The terms are positive for a_0, a_2, a_4, ... (even terms) , while the odd terms a_1, a_3, a_5, ... are negative. (so most likely there should be a factor   (-1) in the common ratio.

We can then represent it with the following summation expression:

\sum^{\infty}_0 {(-3)^n} given that each of its first four terms are:

a_0=(-3)^0=1\\a_1=(-3)^1=-3\\a_2=(-3)^2=(-3)*(-3)=9\\a_3=(-3)^3=(-3)*(-3)*(-3)=-27

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he equation of any horizontal line is

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If θ is an angle in standard position that terminates in Quadrant III such that tanθ = 5/12, then sinθ/2 = _____
babunello [35]

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sin θ/2=5√26/26=0.196

Step-by-step explanation:

θ ∈(π,3π/2)

such that

θ/2 ∈(π/2,3π/4)

As a result,

0<sin θ/2<1, and

-1<cos θ/2<0

tan θ/2=sin  θ/2/cos θ/2

such that

tan θ/2<0

Let

t=tan θ/2

t<0

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2 tan θ/2/1-(tanθ /2)^2  = tanθ

2t/1-t^2=5/12

24t=5 - 5t^2

Solve this quadratic equation for t :

t1=1/5 and

t2= -5

Discard t1 because t is not smaller than 0

Let s= sin θ/2

0<s<1.

By the definition of tangents.

tan θ/2= sin θ/2/ cos θ/2

Apply the Pythagorean Algorithm to express the cosine of θ/2 in terms of s. Note the cos θ/2 is expected to be smaller than zero.

cos θ/2 = -√1-(sin θ/2)^2 = - √1-s^2

Solve for s.

s/-√1-s^2 = -5

s^2=25(1-s^2)

s=√25/26 = 5√26/26

Therefore

sin θ/2=5√26/26=0.196....

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